\newproblem{lay:2_2_7}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.2.7}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Let $A=\begin{pmatrix}1 & 2\\5 & 12\end{pmatrix}$, $\mathbf{b}_1=\begin{pmatrix}-1\\3\end{pmatrix}$, $\mathbf{b}_2=\begin{pmatrix}1\\-5\end{pmatrix}$, 
	$\mathbf{b}_3=\begin{pmatrix}2\\6\end{pmatrix}$, and $\mathbf{b}_4=\begin{pmatrix}3\\5\end{pmatrix}$.
	\begin{enumerate}[a.]
		\item Find $A^{-1}$ and use it to solve the four equations $A\mathbf{x}=\mathbf{b}_1$, $A\mathbf{x}=\mathbf{b}_2$, $A\mathbf{x}=\mathbf{b}_3$, $A\mathbf{x}=\mathbf{b}_4$.
		\item The four equations in part (a) can be solved by the \textit{same} set of row operations, since the coefficients matrix is the same in each case. Solve the four
		      equations in part (a) by reducing the augmented matrix $\left(\begin{array}{r|rrrr} A & \mathbf{b}_1 & \mathbf{b}_2 & \mathbf{b}_3 & \mathbf{b}_4\end{array}\right)$.
	\end{enumerate}
}{
  % Solution
	\begin{enumerate}[a.]
		\item To find $A^{-1}$ we will apply row operations on the augmented matrix exploiting that
		      $\left(\begin{array}{r|r} A & I\end{array}\right) \sim \left(\begin{array}{r|r} I & A^{-1}\end{array}\right)$. \\
		      \begin{center}
						$\left(\begin{array}{rr|rr} 1 & 2 & 1 & 0 \\ 5 & 12 & 0 & 1\end{array}\right) \sim
						 \left(\begin{array}{rr|rr} 1 & 0 & 6 & -1 \\ 0 & 1 & -\frac{5}{2} & \frac{1}{2}\end{array}\right)$
					\end{center}
					Now we use this inverse matrix to solve the linear equations
		      \begin{center}
						$\left(\begin{array}{rr} 6 & -1 \\ -\frac{5}{2} & \frac{1}{2}\end{array}\right)\begin{pmatrix}-1\\3\end{pmatrix}=\begin{pmatrix}-9\\4\end{pmatrix}$\\
						$\left(\begin{array}{rr} 6 & -1 \\ -\frac{5}{2} & \frac{1}{2}\end{array}\right)\begin{pmatrix}1\\-5\end{pmatrix}=\begin{pmatrix}11\\-5\end{pmatrix}$\\
						$\left(\begin{array}{rr} 6 & -1 \\ -\frac{5}{2} & \frac{1}{2}\end{array}\right)\begin{pmatrix}2\\6\end{pmatrix}=\begin{pmatrix}6\\-2\end{pmatrix}$\\
						$\left(\begin{array}{rr} 6 & -1 \\ -\frac{5}{2} & \frac{1}{2}\end{array}\right)\begin{pmatrix}3\\5\end{pmatrix}=\begin{pmatrix}13\\-5\end{pmatrix}$\\
					\end{center}
		\item Now, we will apply row operations on the augmented matrix suggested by the problem
		      \begin{center}
						$\left(\begin{array}{rr|rrrr} 1 & 2 & -1 & 1 & 2 & 3\\ 5 & 12 & 3 & -5 & 6 &5\end{array}\right) \sim
						 \left(\begin{array}{rr|rrrr} 1 & 0 & -9 & 11& 6 &13\\ 0 &  1 & 4 & -5 & -2&-5\end{array}\right)$
					\end{center}
	\end{enumerate}
}
\useproblem{lay:2_2_7}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
